SOLUTION: let n be the largest integer less than 10000 that leaves remainder 1 when divided 2,3,4,5,6,7 and 8.the sum of the digits of n is

Any integer which is 1 more than a multiple of all those
will leave remainder 1 when divided by any of those.  Any 
integer which is a multiple of all those is a multiple of 
their least common multiple.

2,3,4,5,6,7,8 = 2,3,2x2,5,2x3,7,2x2x2

So the LCM must have 3 factors of 2, 1 factor of 3, 1 factor 
of 5, and 1 factor of 7. So the LCM = 2x2x2x3x5x7 = 840, so 
any number of the form 840k+1 will leave remainder 1 when 
divided by 2,3,4,5,6,7 or 8.

We want the largest such integer less than 10000

840k+1 < 10000
  840k < 9999
     k < 9999/840
     k < 11.90357143...

So the largest integer k can be is 11

840k+1 = 840(11)+1 = 9241 = n

The sum of the digits of n is 9+2+4+1 = 16

Edwin